Problem: Simplify and expand the following expression: $ \dfrac{t}{t + 1}+\dfrac{t}{3t + 6} $
In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(t + 1)(3t + 6)$ Multiply the first term by $\dfrac{3t + 6}{3t + 6}$ $ \begin{align*} \dfrac{t}{t + 1} \times \dfrac{3t + 6}{3t + 6} & = \dfrac{(t)(3t + 6)}{(t + 1)(3t + 6)} \\ & = \dfrac{3t^2 + 6t}{(t + 1)(3t + 6)}\end{align*} $ Multiply the second term by $\dfrac{t + 1}{t + 1}$ $ \begin{align*} \dfrac{t}{3t + 6} \times \dfrac{t + 1}{t + 1} & = \dfrac{(t)(t + 1)}{(3t + 6)(t + 1)} \\ & = \dfrac{t^2 + t}{(3t + 6)(t + 1)}\end{align*} $ Now we have: $ = \dfrac{3t^2 + 6t}{(t + 1)(3t + 6)} + \dfrac{t^2 + t}{(3t + 6)(t + 1)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{3t^2 + 6t + t^2 + t}{(t + 1)(3t + 6)} $ $ = \dfrac{4t^2 + 7t}{(t + 1)(3t + 6)}$ Expand the denominator: $ = \dfrac{4t^2 + 7t}{3t^2 + 9t + 6}$